How do you solve #\frac { w - 1} { 5} = \frac { w + 2} { 2}#?

Question

1203 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-02-23T22:52:54

See the entire solution process below:

First, multiply each side of the equation by #color(red)(10)# to eliminate the fractions while keeping the equation balanced:

#color(red)(10) xx (w - 1)/5 = color(red)(10) xx (w + 2)/2#

#cancel(color(red)(10))2 xx (w - 1)/color(red)(cancel(color(black)(5))) = cancel(color(red)(10))5 xx (w + 2)/color(red)(cancel(color(black)(2)))#

#2(w - 1) = 5(w + 2)#

Next, eliminate the parenthesis on each side of the equation by multiplying each term within the parenthesis by the term outside the parenthesis:

#(2 xx w) - (2 xx 1) = (5 xx w) + (5 xx 2)#

#2w - 2 = 5w + 10#

Then, subtract #color(red)(2w)# and #color(blue)(10)# from each side of the equation to isolate the #w# term while keeping the equation balanced:

#2w - 2 - color(red)(2w) - color(blue)(10) = 5w + 10 - color(red)(2w) - color(blue)(10)#

#2w - color(red)(2w) - 2 - color(blue)(10) = 5w - color(red)(2w) + 10 - color(blue)(10)#

#0 - 12 = 3w + 0#

#-12 = 3w#

Now, divide each side of the equation by #color(red)(3)# to solve for #w# while keeping the equation balanced:

#(-12)/color(red)(3) = (3w)/color(red)(3)#

#-4 = (color(red)(cancel(color(black)(3)))w)/cancel(color(red)(3))#

#-4 = w#

#w = -4#